high school Area Between Curves

Area Between Curves

The area trapped between two curves is simpler than it looks. You take the upper curve, subtract the lower curve, and integrate that difference — the result is exactly the area between them. So the real work is finding where the two curves meet: those intersection points become the two limits of the integral, and then you just check which curve sits on top. Drag the right end b below and the region between y=x and y=x² grows, and its area fills up to 1/6.

Subtract the Lower Curve from the Upper

Right end b1

👀 See It

①The area between two curves is the integral of (upper − lower)

②On [0,1], y=x is upper and y=x² is lower

③Growing b to 1 fills the area to 1/6

④Slider b is 0~1 (between intersections). Crossing after x>1 is text

The Area Formula

Area between two curves

S = ∫_a^b (f(x) − g(x)) dx (f ≥ g)

a, b are the x-coordinates of the intersections; f is the upper, g the lower curve

Find the Intersections First

Integration limits = intersections

f(x) = g(x)

Solving the two equations gives the x-values that become the limits a and b

Compute It Directly

Example 1

Find the area enclosed by y=x and y=x².

Find the intersections — solve the two equations.

x = x² ⇒ x = 0, 1

On [0,1] the line x is upper, so integrate (x − x²).

S = ∫₀¹ (x − x²) dx = [x²2 − x³3]₀¹ = 16

▸ 1/6

Plug one point in the interval to check which curve is upper and avoid sign errors.

Example 2

Find the area enclosed by y=x² and the line y=2x.

Find the intersections.

x² = 2x ⇒ x = 0, 2

On [0,2] the line 2x is upper, so integrate (2x − x²).

S = ∫₀² (2x − x²) dx = [x² − x³3]₀² = 43

▸ 4/3

For a curve and a line, the (upper − lower) principle is exactly the same.

Wrap-up

Core strategy

S = ∫_a^b (upper − lower) dx, a·b are intersections

Set the limits by the intersections, then integrate upper minus lower

2022 provincial mock exam Math type, adapted

What is the area enclosed by y=x²−2x and the x-axis?

①2/3

②1

③4/3

④2

⑤8/3

▸ ③ 4/3

Find the intersections with the x-axis (y=0).

x² − 2x = 0 ⇒ x = 0, 2

On [0,2] the curve is below the axis, so integrate (0 − (x²−2x)).

S = ∫₀² (2x − x²) dx = 43

🎯 Exam Points

①Area = ∫(upper − lower)dx

②The limits must be the intersections

③If the curve is below the axis, integrate (0 − curve) = −curve

④If upper and lower swap, split the interval

⑤You may write it with an absolute value, but check the sign on each piece. The basics are on Math II integral applications

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Definite Integrals & Series

Applications of Definite Integrals

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